Forum: Ruby can it be shorter?

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Dorren (Guest)
on 2007-01-25 00:45
(Received via mailing list)
doing access control on rails controller,

------ I have this input --------------
hash = {"index" => "list",
          ["edit", "update"] => "manage_one",
          ["new", "create", "destroy"] => "manage_all"}

------- I want this output ----------
{"index"=>"list",
 "edit"=>"manage_one",
 "update"=>"manage_one",
 "new"=>"manage_all",
 "create"=>"manage_all",
 "destroy"=>"manage_all"
}

------ I have this code ------------
  hash = Hash[*hash.to_a.collect{|x|
                       Array === x[0] ? x[0].zip([x[1]]*x[0].size) : x
                    }.flatten]

----- I want shorter code ------

thanks.
Dorren (Guest)
on 2007-01-25 01:06
(Received via mailing list)
a little clearer than previous one.

  hash = Hash[*hash.inject([]){|arr, (k, v)|
                               arr += Array === k ? k.zip([v]* k.size)
: [k, v]
                             }.flatten]
Vincent F. (Guest)
on 2007-01-25 01:15
(Received via mailing list)
Dorren wrote:
>  "update"=>"manage_one",
> ----- I want shorter code ------
What about this:

h = {}
for k,v in hash
  [k].flatten.each {|l| h[l] = v}
end
hash = h
> => {"new"=>"manage_all", "edit"=>"manage_one",
"destroy"=>"manage_all", "create"=>"manage_all", "index"=>"list",
"update"=>"manage_one"}

  Cheers,

  Vincent
Vincent F. (Guest)
on 2007-01-25 01:22
(Received via mailing list)
Vincent F. wrote:
> h = {}
> for k,v in hash
>   [k].flatten.each {|l| h[l] = v}
> end
> hash = h

  Actually, I've got shorter ;-):

h = {}
for k,v in hash
  [*k].each {|l| h[l] = v}
end
hash = h

  Vince
Dorren (Guest)
on 2007-01-25 01:31
(Received via mailing list)
better, thank tou.

On Jan 24, 6:14 pm, Vincent F. <removed_email_address@domain.invalid>
William J. (Guest)
on 2007-01-25 03:55
(Received via mailing list)
On Jan 24, 5:01 pm, "Dorren" <removed_email_address@domain.invalid> wrote:
> a little clearer than previous one.
>
>   hash = Hash[*hash.inject([]){|arr, (k, v)|
>                                arr += Array === k ? k.zip([v]* k.size)
> : [k, v]
>                              }.flatten]

Hash[ *hash.map{|k,v| k=[*k]; k.zip([v]*k.size) }.flatten ]
Daniel M. (Guest)
on 2007-01-25 04:30
(Received via mailing list)
Vincent F. <removed_email_address@domain.invalid> writes:

>   Actually, I've got shorter ;-):
>
> h = {}
> for k,v in hash
>   [*k].each {|l| h[l] = v}
> end
> hash = h

Well, if we're golfing:

h={};hash.map{|k,v|[*k].map{|t|h[t]=v}};h
Robert K. (Guest)
on 2007-01-25 12:06
(Received via mailing list)
On 25.01.2007 03:29, Daniel M. wrote:
> Well, if we're golfing:
>
> h={};hash.map{|k,v|[*k].map{|t|h[t]=v}};h

Not really shorter but I though there should be at least one solution
with #inject:

 >> hash.inject({}){|h,(k,v)| k.to_a.each {|x| h[x]=v};h}
=> {"new"=>"manage_all", "edit"=>"manage_one", "destroy"=>"manage_all",
"create"=>"manage_all", "index"=>"list", "update"=>"manage_one"}

 >> hash.inject({}){|h,(k,v)| k.each {|x| h[x]=v} rescue h[k]=v;h}
=> {"new"=>"manage_all", "edit"=>"manage_one", "destroy"=>"manage_all",
"create"=>"manage_all", "index"=>"list", "update"=>"manage_one"}

:-)

  robert
Eric H. (Guest)
on 2007-01-25 12:28
(Received via mailing list)
On Jan 24, 2007, at 15:20, Vincent F. wrote:
> for k,v in hash
>   [*k].each {|l| h[l] = v}
> end
> hash = h

but fastest (I assume all entries worked correctly):

$ ruby bm.rb
Rehearsal --------------------------------------------------------
original              10.440000   0.010000  10.450000 ( 10.492738)
original clean        11.480000   0.040000  11.520000 ( 11.632237)
vince                  5.210000   0.010000   5.220000 (  5.264394)
william               12.760000   0.020000  12.780000 ( 12.814230)
daniel                 6.900000   0.010000   6.910000 (  6.924297)
robert inject          6.560000   0.000000   6.560000 (  6.588160)
robert inject rescue   6.080000   0.010000   6.090000 (  6.094140)
---------------------------------------------- total: 59.530000sec

                            user     system      total        real
original              10.450000   0.010000  10.460000 ( 10.467971)
original clean        11.490000   0.030000  11.520000 ( 11.600672)
vince                  5.210000   0.000000   5.210000 (  5.219863)
william               12.820000   0.010000  12.830000 ( 12.839722)
daniel                 6.930000   0.010000   6.940000 (  6.945329)
robert inject          6.580000   0.010000   6.590000 (  6.591095)
robert inject rescue   6.080000   0.000000   6.080000 (  6.117226)
Erik V. (Guest)
on 2007-01-25 13:49
(Received via mailing list)
There's a glitch, which is not yet mentioned...

Consider this old hash: {["A", "B"]=>"C", ["B", "A"]=>"D"}

What's the value of new_hash["A"]? Is it "C", or is it "D"?

That depends on the order on which you build the new hash.
Which depends on the order in which you walk through the old
hash. Which isn't deterministic, AFAIK. Maybe it is
determinstic if you know in which order the old hash was built.
But, given _a_ hash, you simply don't know the order in which
you walk through it, so you don't know the order in which the
new hash will be built, so you don't know what the result is
going to be. Cool...

Adding a sort to the algorithm fixes this.

gegroet,
Erik V. - http://www.erikveen.dds.nl/

----------------------------------------------------------------

 hash.collect do |k,v|
   [[k].flatten, v]
 end.sort.inject({}) do |h,(ks,v)|
   ks.each do |k|
     h[k] = v
   end
   h
 end
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